3.458 \(\int \frac{1}{x^3 (8 c-d x^3)^2 (c+d x^3)^{3/2}} \, dx\)
Optimal. Leaf size=66 \[ -\frac{\sqrt{\frac{d x^3}{c}+1} F_1\left (-\frac{2}{3};2,\frac{3}{2};\frac{1}{3};\frac{d x^3}{8 c},-\frac{d x^3}{c}\right )}{128 c^3 x^2 \sqrt{c+d x^3}} \]
[Out]
-(Sqrt[1 + (d*x^3)/c]*AppellF1[-2/3, 2, 3/2, 1/3, (d*x^3)/(8*c), -((d*x^3)/c)])/(128*c^3*x^2*Sqrt[c + d*x^3])
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Rubi [A] time = 0.0574086, antiderivative size = 66, normalized size of antiderivative = 1.,
number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used =
{511, 510} \[ -\frac{\sqrt{\frac{d x^3}{c}+1} F_1\left (-\frac{2}{3};2,\frac{3}{2};\frac{1}{3};\frac{d x^3}{8 c},-\frac{d x^3}{c}\right )}{128 c^3 x^2 \sqrt{c+d x^3}} \]
Antiderivative was successfully verified.
[In]
Int[1/(x^3*(8*c - d*x^3)^2*(c + d*x^3)^(3/2)),x]
[Out]
-(Sqrt[1 + (d*x^3)/c]*AppellF1[-2/3, 2, 3/2, 1/3, (d*x^3)/(8*c), -((d*x^3)/c)])/(128*c^3*x^2*Sqrt[c + d*x^3])
Rule 511
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && !(IntegerQ[
p] || GtQ[a, 0])
Rule 510
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])
Rubi steps
\begin{align*} \int \frac{1}{x^3 \left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx &=\frac{\sqrt{1+\frac{d x^3}{c}} \int \frac{1}{x^3 \left (8 c-d x^3\right )^2 \left (1+\frac{d x^3}{c}\right )^{3/2}} \, dx}{c \sqrt{c+d x^3}}\\ &=-\frac{\sqrt{1+\frac{d x^3}{c}} F_1\left (-\frac{2}{3};2,\frac{3}{2};\frac{1}{3};\frac{d x^3}{8 c},-\frac{d x^3}{c}\right )}{128 c^3 x^2 \sqrt{c+d x^3}}\\ \end{align*}
Mathematica [B] time = 0.207029, size = 259, normalized size = 3.92 \[ \frac{\frac{64 c \left (-\frac{19648 c^2 d x^3 F_1\left (\frac{1}{3};\frac{1}{2},1;\frac{4}{3};-\frac{d x^3}{c},\frac{d x^3}{8 c}\right )}{3 d x^3 \left (F_1\left (\frac{4}{3};\frac{1}{2},2;\frac{7}{3};-\frac{d x^3}{c},\frac{d x^3}{8 c}\right )-4 F_1\left (\frac{4}{3};\frac{3}{2},1;\frac{7}{3};-\frac{d x^3}{c},\frac{d x^3}{8 c}\right )\right )+32 c F_1\left (\frac{1}{3};\frac{1}{2},1;\frac{4}{3};-\frac{d x^3}{c},\frac{d x^3}{8 c}\right )}-648 c^2-1249 c d x^3+167 d^2 x^6\right )}{8 c-d x^3}+167 d^2 x^6 \sqrt{\frac{d x^3}{c}+1} F_1\left (\frac{4}{3};\frac{1}{2},1;\frac{7}{3};-\frac{d x^3}{c},\frac{d x^3}{8 c}\right )}{663552 c^5 x^2 \sqrt{c+d x^3}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[1/(x^3*(8*c - d*x^3)^2*(c + d*x^3)^(3/2)),x]
[Out]
(167*d^2*x^6*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1/2, 1, 7/3, -((d*x^3)/c), (d*x^3)/(8*c)] + (64*c*(-648*c^2 - 1
249*c*d*x^3 + 167*d^2*x^6 - (19648*c^2*d*x^3*AppellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), (d*x^3)/(8*c)])/(32*c*Ap
pellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), (d*x^3)/(8*c)] + 3*d*x^3*(AppellF1[4/3, 1/2, 2, 7/3, -((d*x^3)/c), (d*x
^3)/(8*c)] - 4*AppellF1[4/3, 3/2, 1, 7/3, -((d*x^3)/c), (d*x^3)/(8*c)]))))/(8*c - d*x^3))/(663552*c^5*x^2*Sqrt
[c + d*x^3])
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Maple [C] time = 0.013, size = 1805, normalized size = 27.4 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/x^3/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x)
[Out]
1/64/c^2*(-2/3*d/c^2*x/((x^3+1/d*c)*d)^(1/2)-1/2/c^2*(d*x^3+c)^(1/2)/x^2+7/18*I/c^2*3^(1/2)*(-d^2*c)^(1/3)*(I*
(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2)*((x-1/d*(-d^2*c)^(1/3)
)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)*(-I*(x+1/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-
d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)
-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^
(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)))-1/64*d/c^2*(-2/27/c^2*x/((x^3+1/d*c)*d)^(1/2)+2/81*I/c^2*3^(1/2
)/d*(-d^2*c)^(1/3)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2)*
((x-1/d*(-d^2*c)^(1/3))/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)*(-I*(x+1/2/d*(-d^2*c)^(1
/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(
x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),(I*3^(1/2)/d*(-d^2*c)^(
1/3)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2))+1/243*I/c^2/d^3*2^(1/2)*sum(1/_alpha^2*(-d
^2*c)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)*(d*(x-1/d*(-d^
2*c)^(1/3))/(-3*(-d^2*c)^(1/3)+I*3^(1/2)*(-d^2*c)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-d^2*c)^(1/3)+(
-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-d^2*c)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-d^2*c)^(2
/3)+2*_alpha^2*d^2-(-d^2*c)^(1/3)*_alpha*d-(-d^2*c)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1
/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),-1/18/d*(2*I*(-d^2*c)^(1/3)*3^(1/2)*_alpha^2*d-
I*(-d^2*c)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-d^2*c)^(2/3)*_alpha-3*c*d)/c,(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3
/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*d-8*c)))+1/8*d/c*(2/243/c^3*x/(
(x^3+1/d*c)*d)^(1/2)-1/1944/c^3*x*(d*x^3+c)^(1/2)/(d*x^3-8*c)-5/1944*I/c^3*3^(1/2)/d*(-d^2*c)^(1/3)*(I*(x+1/2/
d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2)*((x-1/d*(-d^2*c)^(1/3))/(-3/2
/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)*(-I*(x+1/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^
(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*
3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^(1/3)+1
/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2))-1/972*I/c^3/d^3*2^(1/2)*sum(1/_alpha^2*(-d^2*c)^(1/3)*(1/2*I*d*(2*x+1/d
*(-I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)*(d*(x-1/d*(-d^2*c)^(1/3))/(-3*(-d^2*c)^(1/3
)+I*3^(1/2)*(-d^2*c)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3
))^(1/2)/(d*x^3+c)^(1/2)*(I*(-d^2*c)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-d^2*c)^(2/3)+2*_alpha^2*d^2-(-d^2*c)^(
1/3)*_alpha*d-(-d^2*c)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)
)*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),-1/18/d*(2*I*(-d^2*c)^(1/3)*3^(1/2)*_alpha^2*d-I*(-d^2*c)^(2/3)*3^(1/2)*_alp
ha+I*3^(1/2)*c*d-3*(-d^2*c)^(2/3)*_alpha-3*c*d)/c,(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(
1/2)/d*(-d^2*c)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*d-8*c)))
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (d x^{3} + c\right )}^{\frac{3}{2}}{\left (d x^{3} - 8 \, c\right )}^{2} x^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x^3/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="maxima")
[Out]
integrate(1/((d*x^3 + c)^(3/2)*(d*x^3 - 8*c)^2*x^3), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d x^{3} + c}}{d^{4} x^{15} - 14 \, c d^{3} x^{12} + 33 \, c^{2} d^{2} x^{9} + 112 \, c^{3} d x^{6} + 64 \, c^{4} x^{3}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x^3/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="fricas")
[Out]
integral(sqrt(d*x^3 + c)/(d^4*x^15 - 14*c*d^3*x^12 + 33*c^2*d^2*x^9 + 112*c^3*d*x^6 + 64*c^4*x^3), x)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x**3/(-d*x**3+8*c)**2/(d*x**3+c)**(3/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (d x^{3} + c\right )}^{\frac{3}{2}}{\left (d x^{3} - 8 \, c\right )}^{2} x^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x^3/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="giac")
[Out]
integrate(1/((d*x^3 + c)^(3/2)*(d*x^3 - 8*c)^2*x^3), x)